Two dielectric slabs having dielectric constant ' $\mathrm{K}_1$ ' and ' $\mathrm{K}_2$ ' of thickness '…

Two dielectric slabs having dielectric constant ' $\mathrm{K}_1$ ' and ' $\mathrm{K}_2$ ' of thickness ' $\frac{\mathrm{d}}{4}$ and $\frac{3 \mathrm{~d}}{4}$ are inserted between the plates as shown in figure. The net capacitance between $A$ and $B$ is $\left[\varepsilon_0\right.$ is permittivity of free space]
  1. $\frac{2 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}\left[\frac{\mathrm{K}_1 \mathrm{~K}_2}{3 \mathrm{~K}_1+\mathrm{K}_2}\right]$
  2. $\frac{3 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}\left[\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right]$
  3. $\frac{3 \mathrm{A}\varepsilon_0}{2 \mathrm{~d}}\left[\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right]$
  4. $\frac{4 A \varepsilon_0}{d}\left[\frac{K_1 K_2}{3 K_1+K_2}\right]$

Solution

Capacity of $1^{\text {st }}$ Capacitor, $C_1=\frac{K_1 \varepsilon_0 A}{d / 4}=\frac{4 K_1 \varepsilon_0 A}{d}$ Capacity of $2^{\text {nd }}$ Capacitor, $\mathrm{C}_2=\frac{\mathrm{K}_2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d} / 4}=\frac{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}$ Equivalent capacitance $\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}$ $\begin{aligned} \frac{1}{\mathrm{C}_1} & =\frac{\mathrm{d}}{4 \mathrm{~K}_1 \varepsilon_0 \mathrm{~A}} ; \frac{1}{\mathrm{C}_2}=\frac{3 \mathrm{~d}}{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}} \\ \frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \mathrm{~K}_1 \varepsilon_0 \mathrm{~A}}+\frac{3 \mathrm{~d}}{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}} \\ \frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \varepsilon_0 \mathrm{~A}}\left[\frac{1}{\mathrm{~K}_1}+\frac{3}{\mathrm{~K}_2}\right] \\ \frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \varepsilon_0 \mathrm{~A}}\left[\frac{\mathrm{K}_2+3 \mathrm{~K}_1}{\mathrm{~K}_1 \mathrm{~K}_2}\right] \\ \therefore \quad \mathrm{C} & =\frac{4 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left[\frac{\mathrm{K}_1 \mathrm{~K}_2}{3 \mathrm{~K}_1+\mathrm{K}_2}\right] \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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